A person makes a quantity of freed tea by mixing 710 g of hot tea (essentially w

Question

A person makes a quantity of freed tea by mixing 710 g of hot tea (essentially water) with an equal mass of ice at its melting point. Assume the mixture has negligible energy exchanges with its environment, (a) If the tea's initial temperature is T_f 82degree C, when thermal equilibrium is reached what are the mixture's temperature T_f and (b) the remaining mass mf of ice? If T, - 58degreeC, when thermal equilibrium t is reached what are (c) T_v and (d) mf> The specific heat of water is 4186 J/kg-K. The latent heat of fusion is 333 kJ/kg.

Explanation / Answer

heat lost by hot tea in melting ice

Q loss = M*cw*dT = 0.71*4186*(82-0) = 243708.92 J

heat required by ice Qgain= M*L = 0.701*333000 = 233433 J

Qgain < Qloss

all the ice melts

M*Cw*dt = M*L

0.710*4186*(82-t) = 0.71*333000

t = 2.45 degrees <-----------answer

mf = 0 kg    <-----------answer

+++++

(c)

heat lost by hot tea in melting ice

Q loss = M*cw*dT = 0.71*4186*(58-0) = 172379.48 J

heat required by ice Qgain= M*L = 0.701*333000 = 233433 J

Qgain > Qloss

all the ice will not melts

ice and tea are in equilibrium

temperature Tf = 0oC   <-----------answer

M*Cw*dt = (M-mf)*L

0.710*4186*(58-0) = (0.71-mf)*333000

mf = 0.192 kg    <-----------answer

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